Lecture 16: Introduction to the Hydrogen
Atom
The material in this lecture covers the following in Atkins.
Section 13.1 The structure of hydrogenic atoms
(a) The separation of internal motion
Lecture on-line
The Hamiltonian of the hydrogen atom (PDF Format)
The Hamiltonian of the hydrogen atom (PowerPoint)
Handout for this lecture
Audio-visuals on-line
Hydrogen atom Hamiltonian and energy levels
(PowerPoint)(Good account
from the Wilson Group,****)
Hydrogen atom Hamiltonian and energy levels (PDF)
(Good account from the
Wilson Group,****)
Slides from the text book (From the CD included in Atkins ,**)
Interactive Hydrogen Orbital Plots (For Mac users only)
( Visualizes all the
angular and radial wavefunctions of the hydrogen atom, *****)
Hamiltonian for
hydrogenic atom
The hydrogenic atom consists of an electron of
charge (-e) amd mass Me moving around a nuclei
of charge Ze and mass M.
Structure of Hydrogenic Atoms
z
-e
θ
(r,φ,θ)
r
Ze
φ
y
x
The Hamiltonian is given by
p2
+ V( x , y , z )
H =
2µ
meM
µ=
me + M
The reduced mass
Structure of Hydrogenic Atoms
p2
H =
+ V( x , y , z )
2µ
z
-e
θ
(r,φ,θ)
eZ
V( x , y , z ) = −
4πε or
r
Ze
φ
Hamiltonian for
hydrogenic atom
y
x
In quantum mechanics
2
eZ
h
2
ˆ
H = {− ∇ −
}
2µ
4 πε or
Structure of Hydrogenic Atoms
2
h
ˆ = {− ∇ 2 − eZ }
H
2µ
4 πε or
Schrödinger eq. for
hydrogenic atom
h2 2
eZ
{− ∇ −
} ψ ( x , y, z ) = E ψ ( x , y, z )
2µ
4 πε or
We shall solve this eqution
in spherical coordinates
Z
rcosΘ
(x,y,z) → (r, Θ,φ )
Θ
r
φ
Y
X
rsin Θ
x = r sin θ cos φ
y = r sin θ sin φ
z = r cos θ
Structure of Hydrogenic Atoms Schrödinger eq. for
With
hydrogenic atom
eZ
h2 2
{− ∇ −
} ψ ( r , φ, θ ) = E ψ ( r , φ, θ )
2µ
4 πε or
and
∇r2
δ2
2 δ
1 2
=[ 2 +
]− 2 2 L
δ r r δr r h
We can write
2 δ2 2 δ
h
1
Ze
2
ˆ
H=− [ 2 +
]+
L −
2
2µ δ r r δr 2µr
4 πε or
Where
L2 = [
δ2
δθ
2
+ cot θ
δ
δθ
+
1
δ2
sin θ δφ
2
2
]
Structure of Hydrogenic Atoms
Schrödinger eq. for
hydrogenic atom
We finally get the Schrödinger equation
h2 δ 2 2 δ
1
Ze
2
{− [ 2 +
]+
}ψ (r, φ, θ) = Eψ (r, φ, θ)
L −
2
2µ δ r r δr 2µr
4 πε or
z
-e
θ
(r,φ,θ)
r
Ze
φ
x
y
Structure of Hydrogenic Atoms
We shall first show
[L2 , H] = 0
Commutation relations
for hydrogenic atom
We have [L2 , H] =
2 δ2 2 δ
h
1 2
Ze
2
[L , −
(
+
)−
]=
L +
2
2
2µ δ r r δr 2µr
4 πεor
h2 δ 2 2 δ
1
Ze
2
2
L2 ] = 0
[L , −
(
+
)−
] + [L ,
2µ δ 2r r δr 4 πεor
2µr 2
0
Since L2 does not
depend on r
0
Since L2 does not
depend on r
and commutes
with itself
Structure of Hydrogenic Atoms
We can write
r
1 2
h2 δ 2 2 δ
Ze
H(r ) = −
L −
[ 2 +
]+
2
2µ δ r r δr 2µr
4 πε or
We can also show that
[H, L z ] = 0
where
L z = −ih
δ
δφ
Commutation relations
for hydrogenic atom
Structure of Hydrogenic Atoms
We have
r
[L z , H(r )] =
Commutation relations
for hydrogenic atom
h2 δ 2 h2 δ
Ze
1 2
[L z , −
−
−
] + [L z ,
L ] = 0
2µ δ 2r rµ δr 4 πεor
2µr 2
0
0
Since L z does not
depend on r
Since L z commutes
with L2
Thus
[H, L z ] = 0
Structure of Hydrogenic Atoms
Commutation relations
for hydrogenic atom
Since
[L z , H] = 0 ; [L2 , H] = 0 ; [L z , L2 ] = 0 ;
We must be able to find commen
eigenfunctions ψ to all three operators
Hψ = Eψ
L2 ψ = h2 (l + 1)lψ; l = 0,1, 2
L z ψ = mhψ ; m = -l, - l + 1, ....,l
Radial equation
Structure of Hydrogenic Atoms
for hydrogenic atom
We have already shown that
2l + 1 (l− | m!| |m|
Pl (cos θ) × exp[imφ ]
ψ (φ, θ) = Yl,m ((φ, θ) =
4 π (l+ | m!|)
are eigenfunctions to L2 and L z
We must also have that
L2R(r)Yl,m (φ, θ) = h 2 (l + 1)lR(r)Yl,m (φ, θ); l = 0, 1, 2
L zR(r)Yl,m (φ, θ) = mhR(r)Yl,m (φ, θ) ; m = -l, - l + 1, ...., l
as L2 and L z do not depend on r
Structure of Hydrogenic Atoms
Radial equation
for hydrogenic atom
We shall now seek solutions of the form
HR(r)Yl,m (φ, θ) = ER(r)Yl,m (φ, θ)
or
h2 δ 2 2 δ
Ze
1
2
{−
[
+
]+
}R(r)Yl,m (φ, θ)
L +
2
2
2µ δ r r δr 2µmr
4 πεor
= ER(r)Yl,m (φ, θ)
or
h2 δ 2 2 δ
Ze
1
{−
[
+
]+
}R(r)Yl,m (φ, θ) + {
L2 }R(r)Yl,m (φ, θ)
2µ δ 2r r δr 4 πεor
2µmr 2
= ER(r)Yl,m (φ, θ)
Structure of Hydrogenic Atoms
Radial equation
for hydrogenic atom
Thus
h2 δ 2 2 δ
1
Ze
{−
[
+
]+
}R(r)Yl,m (φ, θ) + {
L2 }R(r)Yl,m (φ, θ)
2µ δ 2r r δr 4 πεor
2µmr 2
= ER(r)Yl,m (φ, θ)
Now by working on Yl,m (φ, θ )
h2 δ 2 2 δ
h2l(l + 1)
Ze
Yl,m (φ, θ){−
[
+
]+
}R(r)) + Yl,m (φ, θ){
}R(r)
2µ δ 2r r δr 4 πεor
2µmr 2
= ER(r)Yl,m (φ, θ)
Where we have used that L2 is an eigenfunction of Yl,m (φ, θ )
Structure of Hydrogenic Atoms
Radial equation
for hydrogenic atom
h2 δ 2 2 δ
h2l(l + 1)
Ze
Yl,m (φ, θ){−
[
+
]−
}R(r)) + Yl,m (φ, θ){
}R(r)
2
2
2µ δ r r δr 4 πεor
2µmr
= ER(r)Yl,m (φ, θ)
Multiplying next by 1/ Yl,m (φ, θ ) from the right on both sides
Ze
h2 δ 2 2 δ
h 2l(l + 1)
{−
[ 2 +
]−
}R(r)) + {
}R(r) = ER(r)
2
2µ δ r r δr 4 πε or
2µmr
Combining terms
h 2 δ 2R(r) 2 δR(r)
h 2l(l + 1)
Ze
−
+
{ 2 +
) + {−
}R(r) = ER(r)
2
2µ δ r
r δr
4 πε or 2µmr
Structure of Hydrogenic Atoms
Combining terms
We
h 2have
h 2l(l + 1)
δ 2R(r) 2 δR(r)
Ze
− 2 {22 +
+ 2
) + {−
δr
4Ze
πε or h2lµ(mr
2r δR(r)
l + 12)
h2µ δ δR(r)
r
) + {−
− { 2 +
+
2µ δ r
r δr
4 πε or 2µmr 2
Radial equation
for hydrogenic atom
}R(r) = ER(r)
}R(r) = ER(r)
Veff
Coulomb attraction
Correolis or
centrifugal term
Structure of Hydrogenic Atoms
Solutions to radial equation
for hydrogenic atom
We have
Ze
h 2 δ 2R(r) 2 δR(r)
h 2 l(l + 1)
) + {−
}R(r) = ER(r)
− { 2 +
+
2
2µ δ r
r δr
4 πε or 2µmr
(1) The solutions R(r) to our equation must depend on
l and E
(2) It turns our that for R(r) to be normalized
∞
2
∫ R(r)R(r)r dr = 1
0
E can only take certain values given by
E= -
Z2µe 4
32π 2ε o2 h 2n 2
n = 1, 2, 3...
Structure of Hydrogenic Atoms
We have
Solutions to radial equation
for hydrogenic atom
Ze
h 2 δ 2R(r) 2 δR(r)
h 2 l(l + 1)
) + {−
}R(r) = ER(r)
− { 2 +
+
2
2µ δ r
r δr
4 πε or 2µmr
(3) It is further required that l in the
total solution
ψ (r, φ, θ) = R(r)Yl,m (φ, θ)
satisfies l < n - 1
Thus
ψ (r, φ, θ) = R(r)nl Yl,m (φ, θ)
n = 1, 2, 3 ...
l < n-1
m = - l, - l + 1, ....l - 1, l
Structure of Hydrogenic Atoms
Solutions to radial equation
for hydrogenic atom
Introducing
2Zr
ρ=
ao
and
ao =
4 πε oh 2
me e 2
We find that solutions Rnl (r ) to
Ze
h 2 δ 2Rnl (r) 2 δRnl (r)
h 2l(l + 1)
{
) + {−
}Rnl (r) = EnlRnl (r)
−
+
+
2
2
2µ
4 πε or 2µmr
r δr
δ r
are of the form
ρ l

−ρ / 2n
=
(
r
)
N
L
(
r
)
e
nl
nl   n,l
n
Normalization
Polynomial
Exponential
Structure of Hydrogenic Atoms
n
1
l
0
Solutions to radial equation
for hydrogenic atom
R n, l
 Z
2 
 ao 
3/2
1  Z
 
2 2  ao 
e −ρ / 2
3/2
1
( 2 − ρ)e −ρ / 4
2
2
0
2
1  Z
1 R 2, 0 (r ) =
 
4 6  ao 
3/2
ρe −ρ / 4
Structure of Hydrogenic Atoms
n
l
R nl
3
0
1  Z
 
9 3  ao 
3/2
3/2
3
1  Z
1
 
27 6  ao 
3
2
Solutions to radial equation
for hydrogenic atom
1  Z
 
81 30  ao 
1 2 −ρ / 6
(6 − 2ρ + ρ )e
9
1
(4 − ρ)ρe −ρ / 6
3
3/2
ρ2 e − ρ / 6
What you need to know about the hydrogen atom
from the previous lecture
1. You are not expected to be able
You should also be aware of
to solve the Schrödinger equation for
the following commutation
the hydrogen - like atom
relations
ˆ Ψn,l,m ( r , ϕ, θ) = En Ψn,l,m ( r , ϕ, θ)
H
However you should be aware that the
Hamiltonian can
be written in the form
r
h2 δ 2 2 δ
1
2 − Ze
H=− [ 2 +
L
]+
4 πεo r
2µ δ r r δr 2µr 2
where r is the distance between the
hydrogen - like
atom and the electron whereas µ
is the reduced mass.
ˆ 2 ] = 0 and [H, L
ˆ z] = 0
[H, L
What you need to know about the hydrogen atom
from the previous lecture
2. You are not required to memorize
the exact form of the eigenfunctions.
You should recognized
that they can be
written as a product
of a radial part Rnl (r) and
the spherical harmonics
Ylm (ϕ, θ)
ˆ 2 and L
ˆ z]
[eigenfunctions of L
as
Ψn,l,m (r, ϕ, θ) =
R nl (r)Ylm (ϕ, θ)
with the corresponding
energies given by
E= -
Z2me 4
32 π 2 e o2 h 2n 2
It is important that you
remember the possible
quantum numbers for l
and m with respect to
a given n.